Mathematics · May / June 2024 · Paper 31 · Question 3
YMA01
As printed in the paper
Show the answer and mark scheme
3(i)[3]
y
y
x
y
x
10
10
10
10
10
3
x
log
y
10
log
x
10
States
log
N
=
2
t
+
4
or
log
N
=
log
a
+
t
log
b
3(ii)[3]
3
3
3
3
2
t
+
4
2
t
4
=
3
=
3
×
3
or
log
=
4
=
...
N
a
a
or
3
log
=
2
=
...
b
b
3
t
N =
81 9
×
(i)
Condone invisible brackets around coordinates provided the meaning is clear.
Coordinates may be stated away from the graph which is fine provided a graph is drawn.
Note that they may draw a curve so B0B1B1 is possible as the marks are earned independently of
each other.
Note that if they draw on Figure 2 the axis labels are incorrect so maximum score is B0B1B1
B1:
For a straight line with negative gradient anywhere on a set of axes. It must intersect both axes and
not stop at these points. Do not penalise if the line is not perfectly straight and do not be concerned
with the steepness provided there is no intention to draw a curve or a horizontal line. Axes do not
need to be labelled but if they are then the yaxis needs to be log y and the horizontal axis must be
log x (with or without base 10 labelled)
B1:
For an intercept of 6 on the
log
y axis for their graph. Can be labelled as just 6 instead of (0, 6) .
10
The line must pass through (or start/stop at) this point. Condone y intercept labelled as (6, 0)
provided the point is on the positive yaxis. Do not be concerned by the axis labelling for this mark.
B1:
For an intercept of 2 on the
log
x axis for their graph. Can be labelled as just 2 instead of (2, 0).
10
The line must pass through (or start/stop at) this point. Condone x intercept being labelled as (0, 2)
provided the point is on the positive xaxis. Do not be concerned by the axis labelling for this mark.
Note that
with no working is 0 marks
May work in terms of x instead of t which is fine, but the final answer must be in terms of t
A misread writing a different base to 3 (including 10) can score a maximum B0M1A0.
B1:
For a correct equation
log
2
4
which may be left unsimplified or states
N
=
t
+
3
log
log
log
N
=
a
+
t
b
3
3
3
Condone lack of base 3 written but base 10 is B0. May be implied by further work which is not the
final answer.
If they write down an incorrect equation e.g.
log
N
= −
2
t
+
4
as well as
log
N
=
log
3
3
then allow this mark to be scored.
at
M1:
For using the laws of logs correctly and moving from
log N
=
+
to
N =
3
×
3
at
b
3
t +
2
4
If they write
3
3
then M0A0 but if it is ambiguous between a + and × give the benefit of the
doubt.
This may involve using their
log N
=
at
+
b
for two different values for t to find N using logs
3
correctly. They then form two simultaneous equations using
=
t
and proceed to find a value for
N
ab
a or b.
e.g.
t
=
1
log
N
=
6
N
=
729,
t
=
2
log
N
=
8
N
=
6561
3
3
6561
ab
=
729,
ab
2
=
6561
b
=
or
a
=
81
729
e.g.
(0, 81), ( 2, 1)
−
81
=
0
, 1
=
ab−
2
81 or
9
ab
a
=
b
=
Alternatively, forms an equation in a or b and proceeds to find a value for a or b using logs
correctly.
(may be implied by their values for a or b provided no incorrect log work is seen)
If they misread and write a different base this mark can be scored for working correctly in that base.
e.g.
log
=
4
=
10
4
o.e. or e.g.
N =
10
4
×
10 t
2
a
a
10
t
A1:
N =
81 9
×
cso Just stating the values of a and b does not score this mark. Must be in terms of t
We must see
2
t
+
4
2
t
4
log
=
2
+
4
or
=
3
or
N =
3
×
3
before proceeding to the answer with no incorrect
N
t
N
3
log work seen
4
log
N
=
log
a
+
t
log
b
or both
log
a
=
4
a
=
3
o.e. and
log
b
=
2
b
=
3
3
3
3
3
3
t
proceeding to
N =
81 9
×
2
t
+
4
2
t
4
t
Note that e.g.
=
3
=
3
+
3
=
81 9
×
is B1M0A0 (incorrect log work – the values of 81
N
N
and 9 cannot imply the method mark as we see incorrect log work)
f( )
=
8sin cos
+
4cos
2
−
3
x
x
x
x
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